If z1,z2,z3∈C are the vertices of an equilateral triangle whose centroid is z0, then k=1∑3(zk−z0)2 is equal to:
Solutions · 1
Official solution
Since z0=3z1+z2+z3 is the centroid of an equilateral triangle, it is also the circumcentre, so all three points lie on a circle of radius R centred at z0, spaced 120∘ apart:
\[z_k-z_0 = R e^{i(2k\pi/3+\varphi)}, \quad k=1,2,3.\]
\[\sum_{k=1}^3(z_k-z_0)^2 = R^2 e^{2i\varphi}\left(e^{i4\pi/3}+e^{i8\pi/3}+e^{i4\pi}\right)=R^2e^{2i\varphi}\left(\omega^2+\omega+1\right)=0,\]
since 1+ω+ω2=0.